This question concerns boiling point elevation.

Question

Determine the boiling point of a solution formed by dissolving 0.4000 kg of table sugar, C12H22O11, in 1.100 L of water.

Answer 1

To find the boiling point of this solution, we need to know the molality of the solution, which is defined as moles of solute per kg of solvent. We cannot determine kg of solvent without knowing the density, and since we are using water, we will have to assume a density of 1 g / ml.

molar mass C12H22O11 = 342.3 g / mole

moles of C12H22O11 dissolved = 0.4000 kg x 1000 g / kg x 1 mol / 342.3 g = 1.1686 mols

kg of water = 1.100 L H2O x 1000 ml / L =1100 ml x 1 g / ml = 1100 g (assuming a density of 1g/ml) = 1.100 kg

∆T = imk

∆T = change in boiling point = ?

i = van't Hoff factor = 1 for sucrose (table sugar; a non-electrolyte)

m = molality = moles / kg water = 1.1686 mol / 1.100 kg = 1.062 m

K = boiling point elevation constant for water = 0.512 º/m

Solving for ∆T

∆T = (1)(1.0620)(0.512)

∆T = 0.544º

Boiling point of solution = 100º + 0.5440 = 100.5ºC